Endomorphism/Eigenvectors/Linearly independent/Fact/Proof

Proof

We prove the statement by induction on ${}n$. For ${}n=0$, the statement is true. Suppose now that the statement is true for less than ${}n$ vectors. We consider a representation of ${}0$, say

${}a_{1}v_{1}+\cdots +a_{n}v_{n}=0\,.$

We apply ${}\varphi$ to this and get, on one hand,

${}a_{1}\varphi (v_{1})+\cdots +a_{n}\varphi (v_{n})=\lambda _{1}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.$

On the other hand, we multiply the equation with ${}\lambda _{n}$ and get

${}\lambda _{n}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.$

We look at the difference of the two equations, and get

${}(\lambda _{n}-\lambda _{1})a_{1}v_{1}+\cdots +(\lambda _{n}-\lambda _{n-1})a_{n-1}v_{n-1}=0\,.$

By the induction hypothesis, we get for the coefficients ${}(\lambda _{n}-\lambda _{i})a_{i}=0$, ${}i=1,\ldots ,n-1$. Because of ${}\lambda _{n}-\lambda _{i}\neq 0$, we get ${}a_{i}=0$ for ${}i=1,\ldots ,n-1$, and because of ${}v_{n}\neq 0$, we also get ${}a_{n}=0$.

To fact